GATE CSE 2003 | Question: 64

Question

Let S be a stack of size $n \geq1$. Starting with the empty stack, suppose we push the first n natural numbers in sequence, and then perform $n$ pop operations. Assume that Push and Pop operations take $X$ seconds each, and $Y$ seconds elapse between the end of one such stack operation and the start of the next operation. For $m \geq1$, define the stack-life of $m$ as the time elapsed from the end of $Push(m)$ to the start of the pop operation that removes $m$ from S. The average stack-life of an element of this stack is

• A. $n(X+Y)$
• B. $3Y+2X$
• C. $n(X+Y)-X$
• D. $Y+2X$

Comments

@Mahesh karanam $Correct Bro!!$
\[
\left.
\begin{aligned}
\text{Push 1 }   \rightarrow X \\
\text{Pop }  \rightarrow X
\end{aligned}
\right\} Y
\]
$Here$ $for$ $1$ $element$ $avg$ $stack$  $life$ $is$ $Y.$

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my mistake was in understanding the question because i've thought that push take X sec and Pop takes Y sec and it's what given and it's really a blunder

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https://www.youtube.com/live/QjQ4AtJs3wQ?si=uaeSTekWoF942l7G&t=2640

Answer 1

Let us represent stack-life of $i^{th}$ element as $S(i)$. The $i^{th}$ element will be in stack till $(n-i)$ elements are pushed and popped. Plus one more $Y$ for the time interval between the push of $i^{th }$ element and the $i+1^{th}$ element. So,

$S(i) = Y + 2.(n-i)(Y+X) \\= Y + 2.(n-i)Z \\= Y + 2nZ -2iZ$

where $Z = Y+X$

average stack-life will, $A = \Sigma \frac{S(i)}{n}$

$nA = nY + 2.n.n.Z -2.Z.\Sigma i$

$nA = nY + 2.n.n.Z - 2.Z \frac{(n(n+1))}{2}$

$nA = nY + 2.n.n.Z - Z(n.n) - n.Z$

$A = Y + 2.n.Z - (n+1).Z$

$A = Y + (n-1).Z \\= Y +(n-1)(X+Y) \\= n(X+Y) - X$

Correct Answer: $C$

Comments

I have a doubt. when will we consider an element popped? at the start of the pop operation or at the end? because if at the end then we need to add +x to the avg life which woud give option 1

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Kya baat hai ?🙄

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your answer is in question. "time elapsed from the end of push to the start of the pop"

Answer 2

Lifetime of m : End Of Push to Start of Pop

• n - PUSH  will take $X$ seconds each
• n - POP  will take $X$ seconds each
•  Time Between  two PUSH  or two POP or one PUSH one POP is $Y$ second

Consider an example with stack contails 3 elemets as shown in Image below

Option C

Comments

Why pop2 time no included in lifetime of 2.

@pC

Pls explain.

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@parthbkgadoya when we put n=3, it should be 2X+3Y not 3X+2Y.

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Great answer, really helpful

 

Answer 3

For quick answer:

consider $n=1,$

We pushed only element with value $1$ into the stack.

Time after push is $X$

After a break of time $Y$, we can pop the element from the stack.

Time before pop is $X+Y$

Lifetime of element $1=(X+Y)-X=Y$

Now we can easily verify that the answer is option $C$

Comments

@Milan1999 "stack-life of m as the time elapsed from the end of Push(m) to the start of the pop operation that removes m from S."

Stack life = time from the end of push(m) to the start of pop that removes m.

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wah bhai !!! gajab ki soch rakh te hoo. this aproach is really time saving and less mistake prone

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best answer bhai , Thank You

 

Answer 4

Here is my approach:

Stack lifetime of nth element -> Y
For (n-1)th -> 2X+2Y + stack lifetime of nth element = 2X + 3Y
For (n-2)th -> 2X+2Y + stack lifetime of (n-1)th element = 4X + 5Y
..
..
For 1st -> 2(n-1)X + (2n-1)Y

Sum of all life spans= (Σ 2(n-1)X) + (Σ (2n-1)Y) for n = 1 to n

Calculate sum by the above summation from 1 to n, You will get:

Sum = n(n(X+Y)-X)
Therefore Average = Sum/n = n(X+Y)-X .  Hence Option (c)

source @ http://stackoverflow.com/questions/24464942/stack-life-span

Answer 5

Correct answer is Option C

Please go through below image, there I have solved using Option elimination.

https://photos.app.goo.gl/2bZ0QISbgnpQxCGg2

Comments

If you not understand the solution then paste that image into chatgpt for clear understanding,this was the easiest way to solve this problem

Answer 6

The option for this question is C.